CSIR Numerical Questions with Explain

Question :-
Inclusivefitness of an animal can be measured as a sum of direct
fitness and indirect fitness. Imagine you have 10 offsprings. Through dil-
igent parental care. 5 survive to reproduce. You give your life in a heroic
deed to save a total of 5 of your nieces and nephews . What is your inclu-
sive fitness?
Options
1. 15
2. 12.5
3. 7.5
4. 3.75
Answer: 4
Explanation: Inclusive Fitness = Direct Fitness + Indirect Fitness
= (Survival of offspring) x (r for parent-offspring) +
(Survival of non-descendant kin) x (the proper r for each
type of relationship) In this question Offspring survived= 5 and coeffient of
relatedness(r) =0.5, Niece+nephews= 5 and r= 0.25, hence inclusive fitness =
(5×0.5)+ (5×0.25)= 2.5+1.25= 3.75




Question:- In order to estimate population size of a fish species in lake, a re-
searcher captures 100 fish from the lake and marks them with coloured
tags. A week later, the researcher returns to the lake and catches 150 fish
of the same species and finds that 25 of them are previously tagged ones.
Assuming no immigration or emigration occured, the total population
size of the fish species in the lake will be:
Options
1. 17
2. 38
3. 600
4. 860
Answer: 3
Explanation: n1 x n2/ n3
n1=total marked after catch 1
n2=total marked after catch 2
n3=total caught in catch 2 but marked in catch 1
100×150÷25= 600
Option 3 is correct




Question:- An autosomal recessive condition affects 1 newborn in 10,000 in a
random mating population without any disruptive acting force. What is
the approximate expected frequency of carriers in this population?
Options
1. 1 in 1000 newborns
2. 1 in 500 newborns
3. 1 in 100 newborns
4. 1 in 50 newborns
Answer: 4
Explanation: If we assume that the population’s in H-W equilibrium, then
the frequency of individuals with the diseased genotype is the square of the
frequency of the recessive allele. In other words, freq (aa) = q2 . Freq (aa) =
1÷10000 = 0.0001, and the square root of that is 0.01, which is q, the frequency
of the recessive allele. The frequency of the normal allele is p, equal to 1 - q,
so p = 0.99. The frequency of carriers to be 2pq, which is 0.0198 So 1.98%,
or 198 people out of 10000, should be carriers of autosomal disease, if the
population is in H-W. or 1 amongst 50.